SOLVED EXAMPLES: XRD
Q) A XRD experiment was performed with Cu k radiation, using a diffractometer. The first (&
only) Bragg reflection was observed at 2B  121 (crystal with FCC lattice). What are the
indices of this reflection and the interplanar spacing that it corresponds to? What is the next
reflection observed?
A) Radiation with λ(Cu k) = 1.54Ǻ, only one reflection observed from a FCC material when
2B  121

121
 60.5  sin   0.87 , n  2d sin  , n = 1 (chosen in XRD)
2
1.54  2d (0.87) , d111 =0.8846
For a crystal with FCC lattice the first reflection is from the {111} planes
d111 
a
 a = 1.53 Å
3
Rule for refection to occur in FCC: (h k l) unmixed. The next reflection is expected from the {200}
planes.
d111 sin  200
2
2(0.87)


, sin  200 
 1.0045
d 200 sin 111
3
3
Which is not possible  the next higher reflection (from {200} planes) is not observed.
Q) A XRD experiment was performed with Cu k radiation, using a diffractometer on a Cu-Ni
alloy (FCC solid solution); the first Bragg reflection was observed at  B  22.1 . What is the
composition of the alloy?
A) Given: λ =1.54 Ǻ, 111  22.1o , FCC alloy, Cu-Ni form a isomorphous phase diagram (soluble in
the solid state in all proportions)
Assumption: linear interpolation to get the lattice parameter of the alloy is possible.
From the periodic table at the end of the book: aCu = 3.61 Å, aNi = 3.52 Å.
The lattice parameter of the alloy, where x is the atomic percent of Cu:
am = 3.61x + 3.52(1x)
n  2dm sin  , d m 
(e1)
o
o
1.54
 2.048A , am  3d m  3.544 A
2(0.376)
Substituting in (e1) we get x = 26 atom%  the composition is approximately Cu26Ni74.
1
Q) A XRD experiment was performed with Cr k radiation, using a diffractometer on a cubic
crystal; only two Bragg reflections are observed at B  34.6 &65.51 . What is the crystal
structure and the lattice parameter? Show that these are the only two reflections possible.
A) Given: only two reflections are observed on a powder pattern of a cubic crystal at  = 34.06 and
65.51. Cr K radiation is used with λ = 2.29 Ǻ.
Using Bragg’s law: n  2d sin  and the d spacing formula for a cubic crystal dhkl 
a
h2  k 2  l 2
it is seen that h 2  k 2  l 2  Sin 2  ratios of Sin2 are proportional to ratios of h 2  k 2  l 2 .
Reflection

Sin2
Ratio
1
34.06
0.3136
3
2
65.51
0.8281
8
Planes
 DC structure
{111}
{220}
From the ratios in column 4 of the table it can be concluded that the structure is diamond cubic.
From the selection rules of the diamond cubic structure it is seen that the first two reflections are
from {111} and {220} planes.

 a 
n  2d111 sin  , 2.29  2 
 sin(34.06)  a  3.54 A
 3
The selection rule implies that third reflection if present will be from the {311} planes.
d 311 sin  220
8
11 Sin(65.5o )


, sin 311 
 1.067 (as Sin cannot be greater than 1)
d 220 sin 311
11
8
Which is not possible  the next higher reflection (from {311} planes) is not observed.
Q) A XRD experiment was performed with Cu k radiation, using a diffractometer on a FCC
crystal with lattice parameter = 3.61Ǻ. What are the Miller indices of the planes with lowest and
highest Bragg angles?
A) Given: FCC crystal, a =3.61Ǻ, Wavelength of X-rays- λ =1.54Ǻ
As this is the FCC crystal the planes with lowest Bragg angle are the {111} planes.
Using Bragg’s law: n  2d sin  and noting that n = 1 and
2
For a cubic crystal dhkl 
a
h k l
2
2
2


h2  k 2  l 2

 dmin  (Sin)max = 1. The highest
max
possible Bragg angle corresponds to the highest value of Sin; which should be as close to 1 as
possible.
Substituting the value of  and Sin = 1 in Bragg’s law:
1.54 
2a
h l k
2
2
(1) 
2
 3.61 
2
2
2
h2  l 2  k 2  2 
  2.34  h  l  k  21.90
 1.54 
The plane should be now chosen keeping in view the selection rules for an FCC crystal such that
h 2  l 2  k 2  21.90 .
h
k
l
h2  l 2  k 2
4
0
0
16
3
3
1
19
4
2
0
20

4
2
2
24
Not allowed
Q) The first reflection from SiC is found at an angle of 17.2. What is the Bragg angle for the
second reflection. Clue: SiC has a FCC lattice which gives some weak reflections XRD pattern.
A) Given: 1st reflection from SiC at 17.2
The ratio of h2 + k2 + l2 for the DC structure are:
DC 
3
8
11
16
:
:
:
...
1,1,1  2, 2, 0   3,1,1  4, 0, 0 
A comparison of the reflections between FCC and DC structures is:
(111)
(200)
(220)
(311)
(222)
(400)
(331)
FCC







DC







FCC







(SiC)
(weak)
(weak)
But the structure of SiC is FCC lattice with motif consisting of two atoms: Si at (0,0,0) and C at
(1/4,1/4,1/4). If the atom occupying the two positions were the same then the structure would be DC
and reflections from planes like (200) and (222) would not be observed. In the case of SiC the
3
cancellation of these reflections (i.e (200), (222) etc.) is not complete as the scattering factor of
these
atoms
is
not
equal
and
hence
the
reflection
observed
is
weak.
Comment: The SiC structure is a superlattice structure which can be understood in terms of two
interpenetrating FCC lattices  one with origin at (0,0,0) and other with origin at (1/4,1/4,1/4).
Hence, the weak reflections are called superlattice reflections.
Using Bragg’s law:   2d111 sin 111 

2a
3
sin( 17.2) ,   2d 200 sin  200 
2a
sin  200
2
sin  200
2
 200  19.96

sin(17.2)
3
4